Problem 14
Question
Sketch each region and write an iterated integral of a continuous function \(f\) over the region. Use the order \(d y d x\). a\(R\) is the triangular region with vertices \((0,0),(0,2),\) and (1,1).
Step-by-Step Solution
Verified Answer
Based on the given step-by-step solution, write a short answer for setting up the iterated integral for a continuous function \(f(x, y)\) over the triangle with vertices \((0,0)\), \((0,2)\), and \((1,1)\) using the order \(dy \, dx\).
To set up the iterated integral for the continuous function \(f(x, y)\) over the triangular region with vertices \((0,0)\), \((0,2)\), and \((1,1)\) using the order \(dy \, dx\), you first need to find the equation of the line segment connecting the points \((0,2)\) and \((1,1)\), which is given by \(y = -x + 2\). Then, determine the limits of integration for \(y\) and \(x\), which are \(0\) to \(-x + 2\) for \(y\) and \(0\) to \(1\) for \(x\). Finally, write the iterated integral as follows:
\(\int_0^1 \int_0^{-x+2} f(x, y) \, dy \, dx\)
1Step 1: Sketch the Region
To sketch the region, we first plot the points \((0,0)\), \((0,2)\), and \((1,1)\) on the coordinate plane. Then, connect these points to form a triangle. The region contained inside this triangle is the region \(R\).
2Step 2: Find the Equation of the Line Segment
We need to find the equation of the line segment that connects the points \((0,2)\) and \((1,1)\). The slope of this line segment is given by:
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1-2}{1-0} = -1\)
Since the line passes through the point \((1,1)\), we can write the equation as
\(y - 1 = -1(x - 1)\)
Solving for \(y\), we get
\(y = -x + 2\)
3Step 3: Determine the Limits of Integration
Since the triangle has a horizontal base on the x-axis and we are given the order \(dy \, dx\), we first describe \(y\) in terms of \(x\) and then describe the limits for \(x\). The region \(R\) is bound by the x-axis, the y-axis, and the line \(y = -x + 2\).
For a given \(x\) value in the region:
- The lower limit of \(y\) is \(0\) (because the region is bound by the x-axis).
- The upper limit of \(y\) is given by the line \(y = -x + 2\).
For the \(x\) value in the region:
- The lower limit of \(x\) is \(0\) (because the region is bound by the y-axis)
- The upper limit of \(x\) is \(1\) (the x-coordinate of the vertex where the region terminates).
4Step 4: Write the Iterated Integral
Now, we can write the iterated integral for the continuous function \(f\) over the region \(R\) using the order \(dy \, dx\). We have:
\(\int_0^1 \int_0^{-x+2} f(x, y) \, dy \, dx\)
This iterated integral represents the continuous function \(f\) over the triangular region \(R\).
Key Concepts
Triangular Region IntegrationLimits of IntegrationSketching Regions in CalculusFinding the Equation of a Line
Triangular Region Integration
When we talk about triangular region integration, we are dealing with evaluating integrals over a region that is shaped like a triangle. In calculus, such regions often simplify the process of finding the area or volume under surfaces.
To set up an integral over a triangular region, we first need to identify the vertices of the triangle and sketch it on the coordinate plane. This visual representation helps us determine the boundaries of the triangular region. Once the boundaries are known, we can set the limits for our integrals. Typically, for a right-angled triangle aligned with the axes, this involves integrating first with respect to one variable and then the other, corresponding to the triangle's height and base.
For example, if you are given a function f(x, y), and you want to integrate it over a triangle in the xy-plane, you will need an iterated integral with two integral symbols, each assigned to one variable, like dy and dx. The limits for each integral will be determined by the lines forming the sides of the triangle.
To set up an integral over a triangular region, we first need to identify the vertices of the triangle and sketch it on the coordinate plane. This visual representation helps us determine the boundaries of the triangular region. Once the boundaries are known, we can set the limits for our integrals. Typically, for a right-angled triangle aligned with the axes, this involves integrating first with respect to one variable and then the other, corresponding to the triangle's height and base.
For example, if you are given a function f(x, y), and you want to integrate it over a triangle in the xy-plane, you will need an iterated integral with two integral symbols, each assigned to one variable, like dy and dx. The limits for each integral will be determined by the lines forming the sides of the triangle.
Limits of Integration
The limits of integration are a crucial component when setting up an iterated integral. They define the bounds within which the integral is evaluated and depend on the geometry of the region of integration.
In the context of the given triangular region, the limits for y are from the x-axis up to the line that forms one side of the triangle. For x, the limits are from the y-axis to the vertical line corresponding to the other non-hypotenuse side of the triangle. In this example, for any given x, the value of y starts at 0 and goes up to the line y = -x + 2, determining the upper limit of y. For x, the range is from 0 to 1, which are the intercepts of the triangle on the x-axis.
In the context of the given triangular region, the limits for y are from the x-axis up to the line that forms one side of the triangle. For x, the limits are from the y-axis to the vertical line corresponding to the other non-hypotenuse side of the triangle. In this example, for any given x, the value of y starts at 0 and goes up to the line y = -x + 2, determining the upper limit of y. For x, the range is from 0 to 1, which are the intercepts of the triangle on the x-axis.
Sketching Regions in Calculus
The ability to sketch regions in calculus is fundamental when dealing with integration, as well as in understanding the behavior of functions over specific areas. A sketch is a simple, visual tool that effortlessly communicates the nature of the region you are working with.
When sketching, you start by identifying key points, as demonstrated with the vertices of the triangular region in our example. These are plotted on the Cartesian plane. You then connect these dots to form the geometric shape you are analyzing. For a triangle, once the vertices are connected, you will better grasp how the function behaves within that bounded region. This not only assists in setting up integrals but also in visualizing the real-world significance of the mathematical concept.
When sketching, you start by identifying key points, as demonstrated with the vertices of the triangular region in our example. These are plotted on the Cartesian plane. You then connect these dots to form the geometric shape you are analyzing. For a triangle, once the vertices are connected, you will better grasp how the function behaves within that bounded region. This not only assists in setting up integrals but also in visualizing the real-world significance of the mathematical concept.
Finding the Equation of a Line
Understanding how to find the equation of a line is an essential skill in calculus, especially when dealing with integration over planar regions. Lines form the boundaries for many shapes on the coordinate plane, and thus, determine the limits for our integrals.
The slope-intercept form of a line is given by y = mx + b, where m is the slope and b is the y-intercept. To find the equation of a line, you'll need at least two pieces of information: the slope and one point through which the line passes, or two points located on the line. The slope can be calculated by taking the difference in y-coordinates over the difference in x-coordinates between the two points, and once you have the slope, you can substitute the coordinates of a point on the line into the slope-intercept equation to solve for y. With the equation established, it becomes a powerful tool in setting up and solving integrals over areas bounded by that line.
The slope-intercept form of a line is given by y = mx + b, where m is the slope and b is the y-intercept. To find the equation of a line, you'll need at least two pieces of information: the slope and one point through which the line passes, or two points located on the line. The slope can be calculated by taking the difference in y-coordinates over the difference in x-coordinates between the two points, and once you have the slope, you can substitute the coordinates of a point on the line into the slope-intercept equation to solve for y. With the equation established, it becomes a powerful tool in setting up and solving integrals over areas bounded by that line.
Other exercises in this chapter
Problem 14
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